Giải phương trình sau :
cos3x-sinx=1
Giải phương trình sau :
cos3x-sinx=1
Gọi M,m là giá trị lớn nhất và giá trị nhỏ nhất của hàm số y=2sinx + 2cosx-2sin2x. Khi đó M, m có giá trị nào
`y=2sin x +2cos x-2sin 2x`
`<=>y=2(sin x+cos x-2sin x cos x)`
Đặt `sin x +cos x=t` `t in [-\sqrt{2};\sqrt{2}]`
`=>2sin x cos x=t^2-1`
H/s có dạng: `y=2(t-t^2+1)=-2t^2+2t+2`
Xét `y=-2t^2+2t+2` có: `I(1/2;5/2)`
BBT:
`=>y_[mi n]=-2-2\sqrt{2}<=>t=-\sqrt{2}<=>sin x +cos x=-\sqrt{2}`
`<=>sin(x+\pi/4)=-1<=>x+\pi/4=-\pi/2+k2\pi<=>x=-[3\pi]/4+k2\pi` `(k in ZZ)`
`y_[max]=5/2<=>t=1/2<=>sin (x+\pi/2)=\sqrt{2}/4` `(sin \alpha=\sqrt{2}/4)`
`<=>[(x+\pi/2=\alpha+k2\pi),(x+\pi/2=\pi-\alpha+k2\pi):}`
`<=>[(x=\alpha-\pi/2+k2\pi),(x=\pi/2-\alpha+k2\pi):}` `(k in ZZ)`
Giải phương trình sau:
\(\sqrt{2}sinx+2cos^2x-1-sin2x=0\)
\(\Leftrightarrow\sqrt{2}sinx+cos2x-sin2x=0\)
\(\Leftrightarrow\sqrt{2}sinx=sin2x-cos2x\)
\(\Leftrightarrow\sqrt{2}sinx=\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow sinx=sin\left(2x-\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=x+k2\pi\\2x-\dfrac{\pi}{4}=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{5\pi}{12}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
\(\sin^3x+\cos^3x=1-\dfrac{1}{2}\sin2x\)
\(3\tan^2x+4\sin^2x-2\sqrt{3}\tan x-4\sin x+2=0\)
\(8\cos4x.\cos^22x+\sqrt{1-\cos3x}+1=0\)
a.
\(\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-\dfrac{1}{2}sin2x\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-\dfrac{1}{2}sin2x\right)-\left(1-\dfrac{1}{2}sin2x\right)=0\)
\(\Leftrightarrow\left(sinx+cosx-1\right)\left(1-\dfrac{1}{2}sin2x\right)=0\)
\(\Leftrightarrow\left(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)-1\right)\left(1-\dfrac{1}{2}sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\\sin2x=2\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(\left(3tan^2x-2\sqrt{3}tanx+1\right)+\left(4sin^2x-4sinx+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{3}tanx-1\right)^2+\left(2sinx-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}tanx=\dfrac{1}{\sqrt{3}}\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)
c.
\(\Leftrightarrow4cos4x\left(1+cos4x\right)+\sqrt{1-cos3x}+1=0\)
\(\Leftrightarrow4cos^24x+4cos4x+1+\sqrt{1-cos3x}=0\)
\(\Leftrightarrow\left(2cos4x+1\right)^2+\sqrt{1-cos3x}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2cos4x+1=0\\1-cos3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}16cos^4x-16cos^2x+3=0\\4cos^3x-3cosx-1=0\end{matrix}\right.\)
\(\Rightarrow cosx=-\dfrac{1}{2}\)
\(\Rightarrow x=\pm\dfrac{2\pi}{3}+k2\pi\)
Giải phương trình sau :
16(sin^6x+cos^6x-1)=3sin6x
Giải phương trình sau:
cos^4x-sin^4x=|cosx|+|sinx|
\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=\left|cosx\right|+\left|sinx\right|\)
\(\Leftrightarrow cos^2x-sin^2x=\left|cosx\right|+\left|sinx\right|\)
\(\Leftrightarrow\left(\left|cosx\right|-\left|sinx\right|\right)\left(\left|cosx\right|+\left|sinx\right|\right)=\left|cosx\right|+\left|sinx\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|cosx\right|+\left|sinx\right|=0\left(vn\right)\\\left|cosx\right|-\left|sinx\right|=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left|cosx\right|=1+\left|sinx\right|\)
Do \(\left\{{}\begin{matrix}\left|cosx\right|\le1\\\left|sinx\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|cosx\right|\le1+\left|sinx\right|\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left|cosx\right|=1\\\left|sinx\right|=0\end{matrix}\right.\) \(\Leftrightarrow sinx=0\)
\(\Rightarrow x=k\pi\)
Giải pt sau
Giúp em vs ạ
ĐKXĐ: ...
\(tan2x.tanx+sinx\left(1+\dfrac{sinx.sin\dfrac{x}{2}}{cosx.cos\dfrac{x}{2}}\right)=0\)
\(\Leftrightarrow tan2x.tanx+sinx\left(\dfrac{cosx.cos\dfrac{x}{2}+sinx.sin\dfrac{x}{2}}{cosx.cos\dfrac{x}{2}}\right)=0\)
\(\Leftrightarrow tan2x.tanx+tanx.\dfrac{cos\left(x-\dfrac{x}{2}\right)}{cos\dfrac{x}{2}}=0\)
\(\Leftrightarrow tan2x.tanx+tanx=0\)
\(\Leftrightarrow tanx\left(tan2x+1\right)=0\)
\(\Leftrightarrow...\)
Giúp em câu b vs ạ
\(cos4x+2sin^2x=sin\left(2x+\dfrac{\pi}{2}\right)\)
\(\Leftrightarrow2cos^22x-1+1-cos2x=cos2x\)
\(\Leftrightarrow2cos^22x-2cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
Tìm giá trị lớn nhất, giá trị nhỏ nhất của hàm số : y=1-8sin2xcos2x + 2sin42x
\(y=1-2\left(2sinx.cosx\right)^2+2sin^42x=1-2sin^22x+2sin^42x\)
\(y=1-2sin^22x\left(1-sin^22x\right)=1-2sin^22x.cos^22x\)
\(=1-\dfrac{1}{2}\left(2sin2x.cos2x\right)^2=1-\dfrac{1}{2}sin^24x\)
Do \(0\le sin^24x\le1\Rightarrow\dfrac{1}{2}\le y\le1\)
\(y_{min}=\dfrac{1}{2}\) khi \(sin^24x=1\)
\(y_{max}=1\) khi \(sin4x=0\)
Tìm m để phương trình m.cot^2(2x-π/8)=2m+1 có nghiệm
Với \(m=0\) pt vô nghiệm
Với \(m\ne0\Rightarrow cot^2\left(2x-\dfrac{\pi}{8}\right)=\dfrac{2m+1}{m}\)
Do \(cot^2\left(2x-\dfrac{\pi}{8}\right)\ge0\Rightarrow\) pt có nghiệm khi:
\(\dfrac{2m+1}{m}\ge0\Rightarrow\left[{}\begin{matrix}m>0\\m\le-\dfrac{1}{2}\end{matrix}\right.\)